At vero eos et accusamus et iusto odio dignissimos ducimus qui blanditiis praesentium voluptatum deleniti atque corrupti quos dolores et quas molestias excepturi sint occaecati cupiditate non provident, similique sunt in culpa qui officia deserunt mollitia animi, id est laborum et dolorum fuga. Et harum quidem rerum facilis est et expedita distinctio. Nam libero tempore, cum soluta nobis est eligendi optio cumque nihil impedit quo minus id quod maxime placeat facere possimus, omnis voluptas assumenda est, omnis dolor repellendus. Itaque earum rerum hic tenetur a sapiente delectus, ut aut reiciendis voluptatibus maiores alias consequatur aut perferendis doloribus asperiores repellat.
i have plugged the x coordinates of the points in the x of the function and only graph that has the point (0,3) and (5,8) work and made both sides equal .. and that's answer choices A , B and , C so please help me.
http://assets.openstudy.com/updates/attachments/557a2a37e4b0636b8cc2f7da-doshka_syria-1434069633043-screenshot_1.png the second one here
Yes , i have uploaded them in order .. first one A and B and second one is C and D
How? -5 doesn't work if you plug it in the function it doesn't make both sides equal.
this question only concerns the x-values... \[x^2-4x+3=x+3\]
we know 0 would work because \[(0)^2-4(0)+3=0+3 \] \[3=3\] but since there are multiple coordinates, we need to figure out if there is more than one x value that satisfies the equation
there's 2 intersections... for your choices... so there's more than 1 common point
right, and i have done that and only the points (0,3) and (5,8) do satisfie that function
\[(5)^2-4(5)+3=5+3\] \[25-20+3=5+3\] \[5+3=5+3 \rightarrow 8=8\] yeah that's correct...
\[(-5)^2-4(-5)+3=-5+3\] \[25+20+3 \neq -2\] \[48 \neq -2\]
i think it's the second graph because if you only graph f(x) = x+3 it will only give you a linear graph that it's line only matches the second one and the last one but since the last one doesn't satisfie the function it's not correct and it's B
it is the second graph.. the coordinates of (0,3) and(5,8)
the mistake that i have made when i tried to graph it .. i graphed it like this f(x)=x2 − 4x + 3 = x + 3 .. i should have done the left side as one graph and the second one as another graph and see the final image.
oh... you have to split them up like f(x) = x+3 f(x) = x^2-4x+3 and compare the two together to see if there's a common point.
aha , and that's what i did and the end
it helps to make a table like x f(x) = x+3 f(x) =x^2-4x+3 0 3 3 1 4 0
that's how i knew B and those two points match and the others don't
whenever you see the same number... it's a solution to both f(x) =x+3 and f(x) = x^2-4x+3